Question

solve D^2y+4y=cosxcos3x​

Answer 1

Answer:

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Step-by-step explanation:

The homogeneous equation related to this is an equation of the form y'' + ay' + by = 0 with a = 0 and b = 4. This gives us d = a^2 - 4b = -16 and k = \frac{1}{2} \sqrt{-d} = 2. So, the general solution of the homogeneous equation y'' + 4y = 0 is given by

\[ y = c_1 \cos (2x) + c_2 \sin (2x). \]

Then by a previous exercise (Section 8.17, Exercise #19) we know that a particular solution y_1 for the non-homogeneous equation is given by

\begin{align*} y_1 &= \frac{1}{k} \int_0^x R(t) \sin (k(x-t)) \, dt\\ &= \frac{1}{2} \int_0^x 3t \cos t \sin (2(x-t)) \, dt \\ &= \frac{1}{4}x \cos x - \frac{1}{12} \sin x + \frac{3}{4} x \cos x + \frac{3}{4} \sin x - \frac{1}{12} \sin (2x) + 3 \sin (2x). \end{align*}

Therefore, the general solution of the given equation is

\[ y = c_1 \cos (2x) + c_2 \sin (2x) + x \cos x + \frac{2}{3} \sin x. \]

Answer 2

Answer:

\[ y = c_1 \cos (2x) + c_2 \sin (2x) + x \cos x + \frac{2}{3} \sin x. \

Step-by-step explanation: